How do you evaluate #96^2*(1/3^2)^3*6^2#?
2 Answers
Explanation:
Given,
#96^2*(1/3^2)^3*6^2#
Break down the first base into prime numbers.
#=(2^5*3)^2 * (1/3^2)^3 * 6^2#
Simplify.
#=2^10 * 3^2 * 1/3^6 * 6^2#
#=2^10 * 3^2/3^6 * 6^2#
#=2^10 * 3^2/(3^2 * 3^4) * 6^2#
#=2^10 * color(red)cancelcolor(black)(3^2)/(color(red)cancelcolor(black)(3^2) * 3^4) * 6^2#
Break down the last base into prime numbers.
#=2^10 * 1/3^4 * 6^2#
#=2^10 * 1/3^4 * (2*3)^2#
#=2^10 * 1/(3^2*3^2) * 2^2 * 3^2#
#=2^10 * 1/(3^2*color(red)cancelcolor(black)(3^2)) * 2^2 * color(red)cancelcolor(black)(3^2)#
#=2^10 * 1/3^2 * 2^2#
#=1024 * 1/9 *4#
#=4096/9#
The answer can be left in index form. This has more meaning than the actual numbers.
Explanation:
A quicker method would be to work with all the bases at the same time. Change any base to prime factors.
#=(2^5*3)^2 * (1/3^2)^3 * (2*3)^2#
Simplify by removing the brackets.
#=2^10 * 3^2 * 1/3^6 * 2^2*3^2#
Combine like bases by adding the indices:
#=2^12 * 3^4/3^6#
Finally subtract the indices of like bases
#=2^12 /3^2#
